Question

решите уравнения 4 и 6

Answer 1

$4)quad frac{2}{x-1} - frac{1}{x+1} textgreater 3 ; ,; ; ODZ:; xne 1,; ; xne -1\\ frac{2(x+1)-(x-1)-3(x-1)(x+1)}{(x-1)(x+1)} textgreater 0\\ frac{2x+2-x+1-3x^2+3}{(x-1)(x+1)} textgreater 0\\ frac{-3x^2+x+6}{(x-1)(x+1)} textgreater 0; ; to ; ; frac{3x^2-x-6}{(x-1)(x+1)} textless 0\\3x^2-x-6=0; ; to ; ; D=1+4cdot 3cdot 6=73; ,\\x_1= frac{1-sqrt{73}}{6} approx -1,26; ,; ; x_2= frac{1+sqrt{73}}{6}approx 1,59 \\ frac{3(x- frac{1-sqrt{73}}{6} )(x- frac{1+sqrt{73}}{6} )}{(x-1)(x+1)} textless 0$

$+++(-1,26)---(-1)+++(1)---(1,59)+++\\xin (- frac{1-sqrt{73}}{6} ,-1)cup (1, frac{1+sqrt{73}}{6} )$

$6)quad frac{x-2}{x+2} geq frac{2x-3}{4x-1} ; ,; ; ; ; ODZ:; xne -2; ,; xne frac{1}{4}\\ frac{(x-2)(4x-1)-(2x-3)(x+2)}{(x+2)(4x-1)} geq 0\\ frac{4x^2-9x+2-(2x^2+x-6)}{(x+2)(4x-1)} geq 0; ; to ; ; frac{2x^2-10x+8}{(x+2)(4x-1)} geq 0\\ frac{2(x-1)(x-4)}{(x+2)(4x-1)} geq 0\\+++(-2)---(frac{1}{4})+++[; 1, ]---[, 4, ]+++ \\xin (-infty ,-2)cup (frac{1}{4},1, ]cup [, 4,+infty )$